Mathematics — Standard (041) — CBSE Class 10 board question

Given: Radius r = 21 mm, ∠AOB = 120°

Perimeter of the minor segment = Length of arc AB + Length of chord AB

Step 1: Length of arc AB

$$l = \frac{\theta}{360} \times 2\pi r = \frac{120}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{3} \times 132 = 44 \text{ mm}$$

Step 2: Length of chord AB

Draw OM ⊥ AB. Then ∠AOM = 60°, OA = 21 mm.

$$AM = OA \times \sin 60° = 21 \times \frac{\sqrt{3}}{2} = \frac{21\sqrt{3}}{2}$$

$$AB = 2 \times AM = 21\sqrt{3} = 21 \times 1.73 = 36.33 \text{ mm}$$

Step 3: Perimeter of shaded (minor segment) region

$$= 44 + 36.33 = \textbf{80.33 mm}$$

Source: Chapter 11, Section 11.1 – Areas of Sector and Segment of a Circle

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• Perimeter of a segment = arc length + chord length (not area — a common mistake).
• The angle at the centre is 120°, so arc length uses θ/360 × 2πr.
• For chord length, drop a perpendicular from O to AB; this bisects both AB and ∠AOB, giving ∠AOM = 60°. Use sin 60° = √3/2 to find AM, then AB = 2AM.
• Examiners award 1 mark each for arc length, chord length, and final perimeter.